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In a 110 V DC chopper drive using the CLC scheme, the maximum possible value of the accelerating current is 300 A. The lower limit of the current pulsation is 140 A. What is the maximum limit of current pulsation?
Current Limit Control of Chopper (CLC):
In this strategy, the chopper is turned on when the output current i0 equals a preset value I2.
The chopper is kept on till i0 increases to another preset value I1.
The chopper is turned off when i0 equals I1 and is kept off till the current i0 decays to I2.
The current i0 is thus limited to a maximum of I1 and a minimum of I2 as shown in Figure.
The chopping frequency and the pulse width are dependent on the load parameters.
The advantage of current limit control is that the ripple (I1 - I2), in the load current i0 can be adjusted to a small, predetermined value for a wide range of loads.
The maximum possible value of the accelerating current (I2) = 300 A
Current pulsation or current ripple (ΔI) = 140 A
The waveform can be drawn as,
From the waveform,
The maximum limit of current pulsation (I2) = I1 + ΔI = 300 + 140 = 440 A
In the LC circuit shown in the figure initial current through the inductor is zero, initial voltage across the capacitor is 100 V. Switch S is closed at t = 0 sec. The current through the circuit is
In a LC circuit, the current flows in the circuit is given by,
Where, VS is supply voltage in V
C is capacitance in F
L is inductance in H
ω0 is resonant frequency in rad/sec
From the given circuit, VS = 100 V
C = 1 μF
L = 20 mH
In the circuit shown in figure find the circuit turn off time is ______ μs
In voltage commutation circuit
Circuit turn off time
for highly inductive load
tm = RC ln 2 ------- for resistive load
∴ tm = 15 × 8 × 10-6 ln 2 = 83.18 μs
Consider the circuit shown below:
The effective on period of the chopper if V =230 V,Io = 60 A,C = 55 μF,Ton = 800 μs is:
Mode 1: T1 will remain ON and conduct to load and diode TA will remain OFF
Mode 2: TA will be turned ON; the capacitor current will now flow in reverse direction & T1 stops conducting.
Therefore, the effective time period of the chopper is
Commutation time: It is the time taken to disconnect the load from the supply after the main thyristor is turned OFF.
Effective on period
= (0.8 + 0.42) × 10-3
TON’ = 1.22 ms
What value of capacitor (in μF) is required to forces commutate a thyristor with a turn off time of 20 μs with a 96 V battery and a full load current of 100 A.
The size of capacitor enquired for commutation is